# Exercise 14 Wave Motion

```Exercise 4 (Work and Energy) Suggested Solutions
1. (a) (i) Work done by the force = F s
= 50 N x 0.1 m = 5 J
(ii) Work done by the force = increase in KE
5 = &frac12; mv2
=&gt; 5 = &frac12; (0.1) v2 =&gt; v = 10 m s-1
(b) F = ( mv – mu ) / t =&gt;
[(0.1)(10) – (0.1)(-10)] / 0.2 = 10 N
(c) The first statement is false. By Newton’s first law, a body will move with constant velocity if there is no
external force acting on it. Since there is no friction, the disc can therefore move with constant velocity without
a driving force.
The second statement is true. By Newton’s second law, force is equal to rate of change of momentum.
The faster the disc moves toward the wall, the greater the change of momentum. So the average force exerted
on the wall would be greater.
2.
(c) (i) Acceleration = 8 / 5 = 1.6 m s-2
(ii) Stopping distance = area under graph
= &frac12; x 8 x (15-5) = 40 m
(iii) Deceleration = 8 / (15-5) = 0.8 m s-2
Frictional force = ma = 60 (0.8) = 48 N
(d) The sledge is released from the same height in both cases. The sledge acquires equal speeds (or KE) at
B in both cases. So the stopping distance along BC would remain unchanged.
3. (a)
(b) Friction = mg sin  = 1 x 10 sin 30o = 5 N
(c) a = F / m = 2 / 1 = 2 m s-2
(d) Tilt the runway at a suitable angle such that the trolley will move with a uniform speed after given a
sharp push.
(e) (i) By W = Fs = 2  1 = 2 J
The work done on the trolley by the force is 2 J.
(ii) Gain in kinetic energy = work done by the force = 2 J
(iii) By KE =
1 2
mv
2
=&gt; 2 =
1
1v2
2
=&gt; v = 2 m s1
4.(a) Let f be the air resistance.
Work done by air resistance = kinetic energy loss of the car
1
f  50 =
 500  60 2
2
f = 18 000 N
(Alternative method: F = ma,
v2
60 2
f=m
= 500 
= 18 000 N)
2a
2  50
The average air resistance required is 18 000 N.
(b) (i)
Let h be the height of the plane. By the conservation of energy,
KE = PE + work done against resistance
1

Δ mv 2  = mgh + fs
2

1
h
 500  (602  202) = 500  10  h + 1500 
2
sin 30
h = 100 m
(ii) I will use parachute to stop the car.
It is because from (i) such inclined plane should be 100 m high and
173 m across, where 100 m is roughly the height of a 28-storey
building. It is not practical to use this method.
In addition, such inclined plane can only slow down the car, but
not stopping the car.
1M
1A
1M
1A
1A
1A
1A
MC
1-5 A C E D A
6-10 C A A C B
11-15 B B A A C 16-18 D B A
----------------------------------------------------------------------Explanations to MC
1. Loss in PE = gain in KE
mgx = m v2 / 2
x = v2 / 2g = 2.62 / 2 (10) = 0.338 m
2. Work done = increase in KE
= (1/2)(5) (102 - 62) = 160 J
4. (1) is wrong because both spheres have the same acceleration.
For (2), s = (1/2)gt2
For sphere A, 4h = (1/2)g tA2
For sphere B, h = (1/2) g tB2
So tA = 2tB
For (3), when reaching the ground,
KE of A = mg (4h) = 1(g) (4h) = 4gh
KE of B = mg (h) = 2 g (h) = 2 gh
KE of A = 2 KE of B
5. mg (0.3+0.15) = mvB2 / 2 =&gt; vB = 3 m s-1
mg (0.3-0.1) = m vA2/ 2 =&gt; vA = 2 m s -1
6. Minimum power output = mgh / t = 120 (500) (10) / 60 = 10000 W
7. PE = mgh = mg (H-s) where H is the initial height
PE = - mgs + mgH
The graph is a straight line with negative slope -mg, and positive intercept mgH.
8. Work done against friction = Fs
= F v t = 5 x 10 x 4 = 200 J
9. For (1), PE = mgh, obviously the PE is increasing.
For (2), tension = 24 N, mg = 2 x 10 = 20 N, so net force on the block is T - mg = 24 - 20 = 4 N, therefore
the block must be accelerating, i.e. the kinetic energy must be increasing.
(3) is wrong because
average power = (PE gained + KE gained) / time
PE gained = 2 x 10 x 4 = 80 J
acceleration = F/m = (24 - 20) / 2 = 2 m s-2.
speed gained v = u + at = 0 + 2 x 2 = 4 m s-1.
So KE gained = mv2/2 = 2 x 42 / 2 = 16 J.
=&gt; average power = (80+16)/ 2 = 48 W.
Alternative faster method :
Power developed by the machine
= Fs / t = 24 x 4 / 2 = 48 W
10. Useful output power = KE gained / time
= [m v2 / 2] / 4 = [m (u+at)2 / 2] /4
= [2000 (3 x 4)2 / 2] /4 = 36 kW
11. Energy loss = KE loss
1
x 0.04 (3302 – 2082) = 1312 J
2
12. Output power = mgh / t = mgv = 1000 x 10 x 0.1 = 1000 W
13. Total energy = PE + KE = PEmax (when KE = 0 at highest point)
= mgh = M x 10 x 0.1 = M J. So statement (1) is correct.
During swinging down, loss in PE = gain in KE
mgh = (1/2) mv2
v  2 gh independent of mass
So statement (2) is incorrect.
With a pin fixed at C, the bob will rise to the same level as A. So statement (3) is incorrect.
14. Work done against friction = loss in KE
1
m(40) 2
2
1
In case 2, fd 2  m(80) 2
2
In case 1, fd1 
So d1 : d2 = 402 : 802 = 1:4
15. v = u + (-10) t = u – 10t
KE 
1 2 1
2
mv  mu  10t  (This is a parabola)
2
2
In addition, the KE during the upward journey should decrease to zero at the highest point and then increase
as the diver gains speed.
16. Assume initial height = H
After traveling for d meters, the skier reaches a height h. Then
d sin  = H - h
Multiply both sides by mg, we get
m g d sin = mg (H - h)
PE = mgh = mgH – mgd sin
PE = - mg sin d + mgH
Thus a plot of PE versus d is a straight line with slope – mg sin , and y-intercept mgH.
Net work done = work done by net force = increase in KE
(mg sin - f) d = KE - 0
KE = (mg sin -f) d
So a plot of KE versus d is a straight line with positive slope.
Since loss in PE = gain in KE + work done against friction, so the answer is not C, because in option C, the
loss in PE is exactly equal to gain in KE, this corresponds to frictionless slope.
17. Engine force = mg sin  + friction = mg sin 30o + 0.5 mg = mg
Power of car = engine force x velocity = mg v
18. Loss in PE = gain in KE + work done against friction
mgh = (1/2) mv2 + fs
There is no increase in KE because the block travels with constant speed. So
work done against friction = mgh = 1 x 10 x 2 x sin 30o =10 J
Supplementary Exercise for Exercise 4
1. The figure below shows a conveying belt system in which a belt is driven by a motor. Cases, each of mass
20 kg, are placed on the belt and delivered up the inclined plane at a uniform speed of 0.5 m s-1.
(a) Each case is carried along the inclined plane by the belt without sliding. What is the frictional force
acting on each case by the belt ?
(b) Find the rate of potential energy gained by each case when it is delivered up to the top of the inclined
plane.
(c) Suppose the maximum power of the machine is 1000 W, and the efficiency of the conveying belt is 60
%. Calculate the maximum number of cases the belt can carry at the same time at a uniform speed of
0.5 m s-1.
Multiple Choice Questions
1. A car stopped after emergency braking. The skid mark left by the car was 22.3 m long. Assume that the
friction between the road and tyres was 0.65 times that of the weight of the car. Estimate the speed of the
car when it began to skid.
A. 5.38 m s-1
B. 12.0 m s-1
C. 16.2 m s-1
D. 17.0 m s-1
2. A man pulls a 50 kg block up a smooth inclined plane making an angle
30o with the horizontal as shown. The inclined plane is 6 m long and the
block is pulled from the bottom of the inclined plane to the top in 30 s.
Find the average useful output power of the man.
A. 5 W
B. 10 W
C. 50 W
D. 87 W
3. Which of the following physical quantities is/are vectors?
(1) Force
(2) Work done
(3) Kinetic energy
A (1) only.
(2) only.
C (1) and (2) only.
D (1), (2) and (3).
4. In a marathon, the winner and the runner-up have the same mass. Compared with the runner-up, the winner
has more
A kinetic energy.
B potential energy.
1. (a) Frictional force = mg sin 
= 20 (10) (3/6) = 100 N
(b) Time required = s/v = 6 / 0.5 = 12 s
Rate of PE gained = PE gained / time
= 20 (10) (3) / 12 = 50 W
(c) Let the required number be n
n (50)  1000 (60%)
n = 12
MC 1-4
1.
fs 